Question

Prove that 16 divides n⁴+4n²+11,if n is an odd integer.

Answer 1

n is odd integers .
so, Let n = 2k + 1 , k ∈ ℕ

now, n⁴ + 4n² + 11
= (2k +1)² + 4(2k + 1)² + 11
= (4k² + 4k + 1)² + 4(4k² + 4k + 1) + 11
= (16k⁴ + 16k² + 1 + 32k³ + 8k + 8k²) + 16k² + 16k + 4 + 11
= 16k⁴ + 32k³ + 40k² + 24k + 16
= 16[k⁴ + 2k³ + 1] + 40k² + 24k
= 16[k⁴ + 2k² + 1] + (80k² + 48k)/2
= 16[k⁴ + 2k² + 1] + 16k(5k + 3)/2

here, k(5k + 3)/2 is also a integer for all natural number of k
so, assume k(5k + 3)/2 = m

now, n⁴ + 4n² + 11
= 16[k⁴ + 2k² + 1] + 16m
= 16[k⁴ + 2k² + 1 + m]

hence, it is clear that 16 divides n⁴ + 4n² + 11 when n is odd integers.

Answer 2

Acc to the ques, we have to prove that 16 divides n⁴+4n²+11,if n is an odd integer.

Let us begin with the contradiction. i.e. lets take an even integer n = 2x.

n⁴+4n²+11

or, (n²−1)(n²+5)+16

or, (4x²−1)(4x²+5)+16

or, $16x^{4}$ + 16x² + 11

or, 16x + 11 which is clearly not divisible by 16.

Now lets assume n = 2x + 1.

n⁴+4n²+11

or, (n²−1)(n²+5)+16

or, (4x²+4x)(4x²+4x+6)+16

or,8x(x+1)(2x²+2x+3) + 16

since 8k(k+1)=0(mod16).

Therefore, 16 divides n⁴+4n²+11,if n is an odd integer.